Answer:
$2,800 was invested at 9%.
$7,200 was invested at 6%.
Step-by-step explanation:
Usually, you need to assign variables to the unknowns you are looking for. Then follow the statements you are given to write equations. Then solve the equation or system of equations.
What are we being asked? The amount invested at each rate.
Assign variables:
Let x = amount invested at 6%
Let y = amount invested at 9%
Since we have two unknowns, we need two equations.
Now we follow the statements to write equations.
"you invested in $10,000, part at 6% annual interest and the rest at 9% annual interest."
The total investment is $10,000, so the sum of our two investments, each at an interest rate is $10,000.
First equation:
x + y = 10,000
We have dealt with the two amounts that were invested. Now we deal with the interest earned.
x amount invested at 6% yields 6% of x in interest in 1 year.
6% of x as a decimal is 0.06x.
y amount invested at 9% yields 9% of y in interest in 1 year.
9% of y as a decimal is 0.09y.
The total interest earned at the two rates is 0.06x + 0.09y.
We are told the total interest is $684, so that gives us the second equation.
0.06x + 0.09y = 684
We now have a system of two equations in two unknowns.
x + y = 10,000
0.06x + 0.09y = 684
Let's use the substitution method to solve the system of equations.
We solve the first equation for x:
x = 10,000 - y
Now we replace x of the seconds equation by 10,000 - y.
0.06x + 0.09y = 684
0.06(10,000 - y) + 0.09y = 684
Distribute the 0.06.
600 - 0.06y + 0.09y = 684
0.03y + 600 = 684
0.03y = 84
y = 2,800
$2,800 was invested at 9%.
x + y = 10,000
x + 2,800 = 10,000
x = 7,200
$7,200 was invested at 6%.
Check:
Let's see if 6% of $7,200 plus 9% of $2,800 adds up to $684.
0.06(7200) + 0.09(2800) = 432 + 252 = 684
Yes it does, so our answer is correct.
